Why Is 254 Used in the Braking Distance Formula?

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SUMMARY

The braking distance formula, represented as $$ Bd = \frac{V^2}{254 \times f} $$, utilizes the factor 254 as a conversion constant when velocity (V) is measured in kilometers per hour (km/h). This factor is derived from the standard formula $$\frac{V^2}{2 g \times f}$$, where 2g accounts for gravitational acceleration (approximately 19.6 m/s²). The conversion from (km/h)² to (m/s)² necessitates multiplying by (1/3.6)², resulting in the constant 254 in the denominator. Understanding this relationship clarifies the application of the braking distance formula in practical scenarios.

PREREQUISITES
  • Understanding of basic physics concepts, specifically gravitational acceleration (g).
  • Familiarity with units of measurement, particularly the conversion between kilometers per hour and meters per second.
  • Knowledge of the braking distance formula and its components.
  • Basic algebra skills for manipulating equations.
NEXT STEPS
  • Research the derivation of the braking distance formula in physics textbooks.
  • Learn about unit conversions between kilometers per hour and meters per second.
  • Explore the implications of the adherence coefficient (f) in braking scenarios.
  • Study the effects of different speeds on braking distances in real-world applications.
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, traffic safety analysts, and anyone involved in road safety education or vehicle performance analysis.

hugo_faurand
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Hello everyone !
I'm working on my highway code and my book give me the approximate formula and the real formula of the braking distance. Here's the real formula according to the book :

$$ Bd = \frac{V^2}{254 \times f} $$

With :
V : the velocity
f : the adherence coefficient.

But I have on question. Where does this 254 come from ?

Thanks in advance.
Regards

Hugo
 
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That factor is a units conversion when V is measured in km/h instead of the usual m/s.
 
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Doc Al said:
That factor is a units conversion when V is measured in km/h instead of the usual m/s.

It's strange, usually we multiply (or divide) by 3.6 when we want to convert m/s in km/h (km/h in m/s).
 
hugo_faurand said:
It's strange, usually we multiply (or divide) by 3.6 when we want to convert m/s in km/h (km/h in m/s).
Right. That 254 is more than just a units conversion, it also contains part of the standard formula: $$\frac{V^2}{2 g \times f}$$
 
Doc Al said:
Right. That 254 is more than just a units conversion, it also contains part of the standard formula: $$\frac{V^2}{2 g \times f}$$
Sorry but I don't find the right calculus. 2*9.8 = 19.6 So I don't understand how you find this 254.
 
hugo_faurand said:
Sorry but I don't find the right calculus. 2*9.8 = 19.6 So I don't understand how you find this 254.
2*9.8 takes care of the 2g factor. Now include the conversion from (km/h)^2 to (m/s)^2. That requires multiplying by (1/3.6)^2, which gives you a constant of 2*9.8*3.6^2 = 254 in the denominator.
 
Doc Al said:
2*9.8 takes care of the 2g factor. Now include the conversion from (km/h)^2 to (m/s)^2. That requires multiplying by (1/3.6)^2, which gives you a constant of 2*9.8*3.6^2 = 254 in the denominator.

AH I see. Highway code should tells us the unit of V. Thank you ! :)
 
Great answer. I have also been wondering where the 254 comes from. Now its easier to remember which speed to convert with 3.6 or 0.27778 and which speed does not require conversion because the factor is alredy included in either 254 or 3.6 or 0.694 (TeV). Thank you.
 

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